Question: Solve for $x$ and $y$ using elimination. $\begin{align*}3x+y &= -4 \\ 6x-5y &= 8\end{align*}$
Answer: We can eliminate $x$ when its corresponding coefficients are negative inverses. Recalling our knowledge of least common multiples, multiply the top equation by $-2$ and the bottom equation by $1$ $\begin{align*}-6x-2y &= 8\\ 6x-5y &= 8\end{align*}$ Add the top and bottom equations. $-7y = 16$ Divide both sides by $-7$ and reduce as necessary. $y = -\dfrac{16}{7}$ Substitute $-\dfrac{16}{7}$ for $y$ in the top equation. $3x- \dfrac{16}{7} = -4$ $3x-\dfrac{16}{7} = -4$ $3x = -\dfrac{12}{7}$ $x = -\dfrac{4}{7}$ The solution is $\enspace x = -\dfrac{4}{7}, \enspace y = -\dfrac{16}{7}$.